函数 f, g 和 h 定义如下： f : $x \rightarrow \sin(\pi x), x \in \mathbb{R}, 0 \le x \le 2$, g : $x \rightarrow a^{x}, x \in \mathbb{R}, x \ge -1$, 其中 a 为常数且 0 < a < 1, h : $x \rightarrow f(x), x \in \mathbb{R}, m \le x \le n$。

1. Explain why $f^{-1}$ does not exist.

For $f(x) = \sin(\pi x)$ on the domain $0 \le x \le 2$, we can test values, e.g.,
$f(0) = \sin(0) = 0$
$f(1) = \sin(\pi) = 0$

Since $f(0) = f(1)$, the function maps multiple inputs to the same output. Therefore, $f$ is a many-to-one function, not a one-to-one function, meaning $f^{-1}$ does not exist.

2. Determine the range of $f$ and deduce $m$ and $n$ for $h^{-1}$ to exist.

The minimum and maximum values of $f(x) = \sin(\pi x)$ are $-1$ and $1$.
So, the range of $f$ is $[-1, 1]$.

For $h^{-1}$ to exist, $h$ must be a one-to-one function (i.e. strictly increasing or decreasing). To match the range of $f$ exactly, the domain $[m, n]$ must span exactly between a maximum point and a minimum point within $0 \le x \le 2$.

Maximum: $\sin(\pi x) = 1 \implies \pi x = \frac{\pi}{2} \implies x = \frac{1}{2}$
Minimum: $\sin(\pi x) = -1 \implies \pi x = \frac{3\pi}{2} \implies x = \frac{3}{2}$

Thus, $h$ is strictly decreasing and one-to-one on this interval.
We have $m = \frac{1}{2}$ and $n = \frac{3}{2}$.

3. Evaluate $h^{-1}\left(\frac{1}{2}\right)$.

Let $x = h^{-1}\left(\frac{1}{2}\right)$. This implies $h(x) = \frac{1}{2}$.
$\sin(\pi x) = \frac{1}{2}$

The basic angle is $\frac{\pi}{6}$.
Since the domain of $h$ is $\frac{1}{2} \le x \le \frac{3}{2}$, the angle $\pi x$ must be in the interval $\left[\frac{\pi}{2}, \frac{3\pi}{2}\right]$.
The only solution in this interval (second quadrant) is:
$\pi x = \pi - \frac{\pi}{6} = \frac{5\pi}{6}$
$x = \frac{5}{6}$

1. Determine the range of the inner function $h$.

From part (i), the range of $h$ is given to be equal to the range of $f$.
For $f(x) = \sin(\pi x)$ where $0 \le x \le 2$, the curve covers a full cycle. The maximum value is $1$ and the minimum value is $-1$.
$R_h = [-1, 1]$

2. Show that the composite function $gh$ exists by verifying the necessary domain-range condition.

The domain of $g$ is given as $x \ge -1$, which means $D_g = [-1, \infty)$.
For the composite function $gh$ to exist, the range of $h$ must be a subset of the domain of $g$ ($R_h \subseteq D_g$).
Since $R_h = [-1, 1]$ and $[-1, 1] \subseteq [-1, \infty)$, the condition is satisfied.
Hence, the composite function $gh$ exists.

3. Find the range of the composite function $gh$.

The range of $gh$ is the set of values of $g(y)$ for all $y \in R_h$.
Substitute the interval $y \in [-1, 1]$ into $g(y) = a^y$.
Since the base $0 < a < 1$, $g(y)$ is a strictly decreasing exponential function.
Minimum value $= g(1) = a^1 = a$
Maximum value $= g(-1) = a^{-1} = \frac{1}{a}$
Therefore, the range of $gh$ is $\left[a, \frac{1}{a}\right]$.